0 votes
29 views
ago in ElectroChemistry by (490 points)
Determine the corrosion current density of a copper plate in a 0.1 M NaCl solution at 25°C, given that the corrosion potential of the copper plate in the solution is -0.54 V vs. the standard hydrogen electrode (SHE) and the Tafel slope for copper in the solution is 120 mV/decade.

1 Answer

0 votes
ago by (690 points)
To determine the corrosion current density of the copper plate in the 0.1 M NaCl solution, we need to use the Tafel equation. The Tafel equation relates the corrosion potential  E_corr , the Tafel slope  b , and the corrosion current density  i_corr  as follows:E = E_corr - b * log10 i_corr / i Where:E = corrosion potential  -0.54 V vs. SHE E_corr = equilibrium potential of the copper plateb = Tafel slope  120 mV/decade = 0.12 V/decade i_corr = corrosion current density  A/m i = exchange current density  A/m First, we need to find the equilibrium potential  E_corr  of the copper plate in the 0.1 M NaCl solution. The standard reduction potential for copper  Cu/Cu  is +0.34 V vs. SHE. Since the corrosion potential of the copper plate in the solution is -0.54 V vs. SHE, the equilibrium potential  E_corr  is:E_corr = E -  -0.54 V  = 0.34 V + 0.54 V = 0.88 VNow, we can rearrange the Tafel equation to solve for the corrosion current density  i_corr :i_corr = i * 10^ E - E_corr  / b We don't have the value for the exchange current density  i  for the copper plate in the 0.1 M NaCl solution. However, we can estimate it using typical values for copper in similar environments. A common value for i for copper in chloride solutions is around 10^-6 A/m.i  10^-6 A/mNow, we can plug in the values into the equation:i_corr =  10^-6 A/m  * 10^ -0.54 V - 0.88 V  / 0.12 V/decade i_corr =  10^-6 A/m  * 10^-1.42 V / 0.12 V/decade i_corr =  10^-6 A/m  * 10^-11.83 i_corr  1.47 * 10^-18 A/mThe corrosion current density of the copper plate in the 0.1 M NaCl solution at 25C is approximately 1.47 * 10^-18 A/m.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...